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# Is (x+y)(x-y) = even integer?

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Is (x+y)(x-y) = even integer?  [#permalink]

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Updated on: 07 Apr 2015, 04:40
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Is $$(x+y)(x-y) =$$ even integer?

1) $$x^2+y^2 =$$ even

2) $$x+y =$$even

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Originally posted by Harley1980 on 06 Apr 2015, 11:19.
Last edited by Harley1980 on 07 Apr 2015, 04:40, edited 3 times in total.
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Re: Is (x+y)(x-y) = even integer?  [#permalink]

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07 Apr 2015, 02:08
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Harley1980 wrote:
Is $$(x+y)(x-y) =$$ even integer?

1) $$x^2+y^2 =$$ even

2) $$x+y =$$even

So now the question is - what is the actual answer?

Try using both statements together: If you know that $$x + y$$ is an even integer and $$x^2 + y^2$$ is an even integer, is it necessary that $$x^2 - y^2$$ should be an even integer too? Actually, no!

It is easy to show that if x and y are both even/both odd integers, then $$x+y$$ is even, $$x^2 + y^2$$ is even and $$x^2 - y^2$$ is even.

But how do you figure out a case where $$x+y$$ is even, $$x^2 + y^2$$ is even but $$x^2 - y^2$$ is not an even integer?

Let me give you an example: Say $$x = 3+\sqrt{6}$$ and $$y = 3 - \sqrt{6}$$.

Here, $$x + y = 6$$ (even integer)
$$x^2 + y^2 = 30$$ (even integer)
But $$x^2 - y^2 = 12\sqrt{6}$$

What you should think about is: How can you prove with variables that answer is (E)? Perhaps, the form of x and y that I have given as an example can help you!
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Re: Is (x+y)(x-y) = even integer?  [#permalink]

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06 Apr 2015, 14:22
Am i missing something simple?

A)(x+y)(x-y) = x^2+y^2 and we are told that is even... so suff

B) we are told (x+y) is even. even * even = even and odd * even = even . Even if (x-y) = 0, gmat considers 0 as even. So Suff
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Re: Is (x+y)(x-y) = even integer?  [#permalink]

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06 Apr 2015, 14:30
Wofford09 wrote:
Am i missing something simple?

A)(x+y)(x-y) = x^2+y^2 and we are told that is even... so suff

B) we are told (x+y) is even. even * even = even and odd * even = even . Even if (x-y) = 0, gmat considers 0 as even. So Suff

about A) $$(x+y)(x-y)$$ not equal to $$x^2+y^2$$ it is equal to $$x^2-y^2$$

Let's assume that $$x =\frac{5}{3}$$ and $$y = \frac{1}{3}$$, $$x + y = \frac{5}{3}+\frac{1}{3} = \frac{6}{3} = 2$$

but $$x^2+y^2$$ will be equal to $$(\frac{5}{3})^2 + (\frac{1}{3})^2 = \frac{25}{9}+\frac{1}{9} =\frac{26}{9}$$ not even
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Re: Is (x+y)(x-y) = even integer?  [#permalink]

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06 Apr 2015, 20:50
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Harley1980 wrote:
Wofford09 wrote:
Am i missing something simple?

A)(x+y)(x-y) = x^2+y^2 and we are told that is even... so suff

B) we are told (x+y) is even. even * even = even and odd * even = even . Even if (x-y) = 0, gmat considers 0 as even. So Suff

about A) $$(x+y)(x-y)$$ not equal to $$x^2+y^2$$ it is equal to $$x^2-y^2$$

Let's assume that $$x =\frac{5}{3}$$ and $$y = \frac{1}{3}$$, $$x + y = \frac{5}{3}+\frac{1}{3} = \frac{6}{3} = 2$$

but $$x^2+y^2$$ will be equal to $$(\frac{5}{3})^2 + (\frac{1}{3})^2 = \frac{25}{9}+\frac{1}{9} =\frac{26}{9}$$ not even

Using the same logic, think what happens in stmnt 1 when $$x^2 = 5/3$$ and $$y^2 = 1/3$$.
In this case, $$x^2 + y^2 = 2$$ (even) but $$x^2 - y^2 = 4/3$$ (not an even integer)
but if $$x^2 = 12$$ and $$y^2 = 6$$, both $$x^2 + y^2$$ and $$x^2 - y^2$$ are even.
So how can the answer be (A)?
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Re: Is (x+y)(x-y) = even integer?  [#permalink]

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06 Apr 2015, 23:15
VeritasPrepKarishma wrote:
Harley1980 wrote:
Wofford09 wrote:
Am i missing something simple?

A)(x+y)(x-y) = x^2+y^2 and we are told that is even... so suff

B) we are told (x+y) is even. even * even = even and odd * even = even . Even if (x-y) = 0, gmat considers 0 as even. So Suff

about A) $$(x+y)(x-y)$$ not equal to $$x^2+y^2$$ it is equal to $$x^2-y^2$$

Let's assume that $$x =\frac{5}{3}$$ and $$y = \frac{1}{3}$$, $$x + y = \frac{5}{3}+\frac{1}{3} = \frac{6}{3} = 2$$

but $$x^2+y^2$$ will be equal to $$(\frac{5}{3})^2 + (\frac{1}{3})^2 = \frac{25}{9}+\frac{1}{9} =\frac{26}{9}$$ not even

Using the same logic, think what happens in stmnt 1 when $$x^2 = 5/3$$ and $$y^2 = 1/3$$.
In this case, $$x^2 + y^2 = 2$$ (even) but $$x^2 - y^2 = 4/3$$ (not an even integer)
but if $$x^2 = 12$$ and $$y^2 = 6$$, both $$x^2 + y^2$$ and $$x^2 - y^2$$ are even.
So how can the answer be (A)?

Wow, that's amazing. VeritasPrepKarishma, you have an eagle eye )
Thank you.
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Re: Is (x+y)(x-y) = even integer?  [#permalink]

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Updated on: 07 Apr 2015, 00:22
My solution:
For the first option: $$X^2 + Y^2 =$$ even, then if you add or subtract $$2*X*Y$$, you will get an even number aswell, so $$X^2 + Y^2 + 2*X*Y = (X + Y)^2$$ is even and $$(X - Y)^2$$ is even too. Say both of them are $$2*k$$ and $$2*b$$ respectively.
$$\sqrt{2*k}*\sqrt{2*b} = 2*\sqrt{k}*\sqrt{b}$$ which is even, thus #1 is sufficient.
For the second - well, its quite obvious: product of even number with whatever = even, thats true, thus sufficient

C it is then
edit: meant D, which is, sadly, incorrect.

Originally posted by Zhenek on 07 Apr 2015, 00:11.
Last edited by Zhenek on 07 Apr 2015, 00:22, edited 1 time in total.
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Re: Is (x+y)(x-y) = even integer?  [#permalink]

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07 Apr 2015, 00:15
Zhenek wrote:
My solution:
For the first option: $$X^2 + Y^2 =$$ even, then if you add or subtract $$2*X*Y$$, you will get an even number aswell, so $$X^2 + Y^2 + 2*X*Y = (X + Y)^2$$ is even and $$(X - Y)^2$$ is even too. Say both of them are $$2*k$$ and $$2*b$$ respectively.
$$\sqrt{2*k}*\sqrt{2*b} = 2*\sqrt{k}*\sqrt{b}$$ which is even, thus #1 is sufficient.
For the second - well, its quite obvious: product of even number with whatever = even, thats true, thus sufficient

C it is then

Zhenek, this is look as you say that answer D is right: both statements are sufficient by themselves. Am I right?
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Re: Is (x+y)(x-y) = even integer?  [#permalink]

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Updated on: 07 Apr 2015, 02:35
Harley1980 wrote:
Zhenek wrote:
My solution:
For the first option: $$X^2 + Y^2 =$$ even, then if you add or subtract $$2*X*Y$$, you will get an even number aswell, so $$X^2 + Y^2 + 2*X*Y = (X + Y)^2$$ is even and $$(X - Y)^2$$ is even too. Say both of them are $$2*k$$ and $$2*b$$ respectively.
$$\sqrt{2*k}*\sqrt{2*b} = 2*\sqrt{k}*\sqrt{b}$$ which is even, thus #1 is sufficient.
For the second - well, its quite obvious: product of even number with whatever = even, thats true, thus sufficient

C it is then

Zhenek, this is look as you say that answer D is right: both statements are sufficient by themselves. Am I right?

Oh, yea, I guess I meant that indeed, not experienced with the gmat thing yet. Looks like my answer is wrong then, what a bummer. I guess I really need to pay attention to the given info (no information given about X and Y being non-integers, which completely blows my solution from the get-go: $$2*X*Y$$ could be $$2*\sqrt{5}/2*\sqrt{3}/2$$ which is not even remotely even )

Originally posted by Zhenek on 07 Apr 2015, 00:18.
Last edited by Zhenek on 07 Apr 2015, 02:35, edited 2 times in total.
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Re: Is (x+y)(x-y) = even integer?  [#permalink]

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07 Apr 2015, 02:38
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Well then, lets amend my approach. Lets start from get-go.

Looking at 1st option and taking into account the fact that X and Y could be non-integers, we can't really say anything about the answer to the question with just option 1 on its own, I'd just call it insufficient right away after comming up with 2 different examples, one being integer and another - non-integer.
Easiest ones that come into mind are:
1)X = 1, Y = 1: 2 is even, 0 is even
2)X = $$\sqrt{5}/2$$, Y = $$\sqrt{3}/2$$: 2 is even, 1/2 is not even
This makes #1 insufficient on its own.

#2 - same story, take fractions as second example and integers as first example, insufficient on its own.

Now lets look at them together
$$x^2 + y^2 = 2*k$$
$$x + y = 2*m$$ => $$y = 2*m - x$$
$$x^2 + y^2 = (x+y)^2 - 2*x*y = (x+y)^2 - 2*x*(2*m-x) = 4*m^2 -2*x*(2*m - x) = 2*k$$
$$2*k = 4*m^2 - 2*x*(2*m - x)$$ => $$x^2 -2*m*x + 2*m^2 - k = 0$$
$$x = m$$±$$\sqrt{k - m^2}$$
$$y = 2*m - x = m$$∓$$\sqrt{k - m^2}$$
So we found values of X and Y that would match 2 of these equation(both #1 and #2). That being said, k and m are random positive integers thus we can't be sure if the expression under the root is a perfect square or not.
If you input these values into your question, you will get $$x^2 - y^2 = (x-y)*(x+y) =$$±$$4*m*\sqrt{k - m^2}$$ which unfortunatelly doesn't answer our question coz of root's value being uncertain (perfect square, then answer is "even", if it is not perfect square, then answer is "not even")
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Re: Is (x+y)(x-y) = even integer?  [#permalink]

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07 Apr 2015, 04:43
Looks like my first attempt to make a task became a train wreck

VeritasPrepKarishma, Zhenek thanks for your explanations, it is now clear. I've changed answer to E.
Looks like this isn't 600-700 lvl task.
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Is (x+y)(x-y) = even integer?  [#permalink]

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28 Nov 2017, 12:04
I had a hard time with this question.

I would like to better understand what went wrong in my approach:

The question is: $$(X + Y)(X - Y)$$ = Even?
Which I rephrased too: $$X^2 + Y^2 - 2XY$$ = Even?
Hence, when looking at statement 1 and 2, I'm able to asses that X and Y need to either both be Even or Odd, and $$2XY$$ has to be Even. So I just went on to select D.

Thanks,
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Re: Is (x+y)(x-y) = even integer?  [#permalink]

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29 Nov 2017, 03:38
I had a hard time with this question.

I would like to better understand what went wrong in my approach:

The question is: $$(X + Y)(X - Y)$$ = Even?
Which I rephrased too: $$X^2 + Y^2 - 2XY$$ = Even?
Hence, when looking at statement 1 and 2, I'm able to asses that X and Y need to either both be Even or Odd, and $$2XY$$ has to be Even. So I just went on to select D.

Thanks,

Note that $$(X + Y)*(X - Y) = X^2 - Y^2$$
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Re: Is (x+y)(x-y) = even integer?  [#permalink]

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30 Nov 2017, 18:06
OMG

I swear I have not been drinking. Just working too hard! Nonetheless, this is unacceptable of me. Thanks very much Karishma, and forgive my carelessness.

VeritasPrepKarishma wrote:
I had a hard time with this question.

I would like to better understand what went wrong in my approach:

The question is: $$(X + Y)(X - Y)$$ = Even?
Which I rephrased too: $$X^2 + Y^2 - 2XY$$ = Even?
Hence, when looking at statement 1 and 2, I'm able to asses that X and Y need to either both be Even or Odd, and $$2XY$$ has to be Even. So I just went on to select D.

Thanks,

Note that $$(X + Y)*(X - Y) = X^2 - Y^2$$

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November 25th, 2017: CAT 3 - Admit Master (GoGMAT) - 710 (Q: 48, V: 40)
November 27th, 2017: CAT 4 - GMATPrep - 720 (Q: 49, V: 40)

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Re: Is (x+y)(x-y) = even integer?  [#permalink]

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01 Dec 2017, 00:30
OMG

I swear I have not been drinking. Just working too hard! Nonetheless, this is unacceptable of me. Thanks very much Karishma, and forgive my carelessness.

VeritasPrepKarishma wrote:
I had a hard time with this question.

I would like to better understand what went wrong in my approach:

The question is: $$(X + Y)(X - Y)$$ = Even?
Which I rephrased too: $$X^2 + Y^2 - 2XY$$ = Even?
Hence, when looking at statement 1 and 2, I'm able to asses that X and Y need to either both be Even or Odd, and $$2XY$$ has to be Even. So I just went on to select D.

Thanks,

Note that $$(X + Y)*(X - Y) = X^2 - Y^2$$

It is better to make these errors during practice so that you do not make them during the actual exam!!
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Re: Is (x+y)(x-y) = even integer? &nbs [#permalink] 01 Dec 2017, 00:30
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